已知(x+2)²+|y+1|=0,求(2x²-xy)-2(x²-y²+xy)+(2xy-2y²) 的值
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![已知(x+2)²+|y+1|=0,求(2x²-xy)-2(x²-y²+xy)+(2xy-2y²) 的值](/uploads/image/z/1730654-62-4.jpg?t=%E5%B7%B2%E7%9F%A5%28x%2B2%29%26%23178%3B%2B%7Cy%2B1%7C%3D0%2C%E6%B1%82%282x%26%23178%3B-xy%29-2%28x%26%23178%3B-y%26%23178%3B%2Bxy%29%2B%282xy-2y%26%23178%3B%29+%E7%9A%84%E5%80%BC)
已知(x+2)²+|y+1|=0,求(2x²-xy)-2(x²-y²+xy)+(2xy-2y²) 的值
已知(x+2)²+|y+1|=0,求(2x²-xy)-2(x²-y²+xy)+(2xy-2y²) 的值
已知(x+2)²+|y+1|=0,求(2x²-xy)-2(x²-y²+xy)+(2xy-2y²) 的值
(x+2)²+|y+1|=0
则x+2=0 y+1=0
x= -2 y= -1
(2x²-xy)-2(x²-y²+xy)+(2xy-2y²)
=2x²-xy-2x²+2y²-2xy+2xy-2y²
= -xy
= -(-2)×(-1)
= -2
(x+2)²+|y+1|=0
x=-2,y=-1
(2x²-xy)-2(x²-y²+xy)+(2xy-2y²)
= 2x²-xy - 2x²+2y²-2xy+2xy-2y²
=-xy
= -2*(-1)
= 2
-2
由括号里x+2的平方+绝对值里y+1等于零可知x+2=0,y+1=0所以x=-2,y=-1.带入后式可以得到最后结果是0